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The work function of a metal is 2 eV. If a radiation of wave length $3000 A °$ is incident on it. Maximum kinetic energy of emitted electron is ______ x $10^{- 19} J .$

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4.4

3.4

2.5

answer is C.

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Detailed Solution

$E = ψ + KE$

$K = E - ψ$

$K = \frac{hc}{λ} - 2 eV$

$= \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{3 \times 10^{- 7}} - 2 \times 1.6 \times 10^{- 19} J$

$= 3.4 \times 10^{- 19} J$

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